抽象函數(shù)問(wèn)題課件高三數(shù)學(xué)二輪復(fù)習(xí)

2025高考數(shù)學(xué)二輪復(fù)習(xí)抽象函數(shù)問(wèn)題抽象函數(shù)是指沒(méi)有具體地給出解析式,只給出它的一些特征或性質(zhì)的函數(shù);抽象函數(shù)型綜合問(wèn)題,一般通過(guò)對(duì)函數(shù)性質(zhì)的代數(shù)表述,綜合考查學(xué)生對(duì)于數(shù)學(xué)符號(hào)語(yǔ)言的理解和接受能力,考查對(duì)于函數(shù)性質(zhì)的代數(shù)推理和論證能力,考查學(xué)生對(duì)于一般和特殊關(guān)系的認(rèn)識(shí),是考查學(xué)生能力的較好途徑.抽象函數(shù)問(wèn)題既是高考的難點(diǎn),又是近幾年高考的熱點(diǎn).賦值法的應(yīng)用例1(1)(多選題)(2023新高考Ⅰ,11)已知函數(shù)f(x)的定義域?yàn)镽,f(xy)=y2f(x)+x2f(y),則(

)A.f(0)=0B.f(1)=0C.f(x)是偶函數(shù)D.x=0為f(x)的極小值點(diǎn)ABC解析

對(duì)于選項(xiàng)A,令x=0,y=0,f(0)=0,所以A正確;對(duì)于選項(xiàng)B,令x=1,y=1,f(1×1)=12×f(1)+12×f(1)=2f(1),解得f(1)=0,所以B正確;對(duì)于選項(xiàng)C,令x=-1,y=-1,f[(-1)×(-1)]=(-1)2×f(-1)+(-1)2×f(-1)=2f(-1),解得f(-1)=0;再令x=-1,y=x,f[(-1)×x]=x2×f(-1)+(-1)2×f(x),f(-x)=f(x),所以C正確;對(duì)于選項(xiàng)D,用特值法,函數(shù)f(x)=0,為常數(shù)函數(shù),且滿(mǎn)足f(xy)=y2f(x)+x2f(y),而常數(shù)函數(shù)沒(méi)有極值點(diǎn),所以D錯(cuò)誤.故選ABC.(2)(2024遼寧撫順一模)已知定義域?yàn)閧x|x≠0}的函數(shù)f(x)滿(mǎn)足對(duì)任意x≠0,y≠0,有f(x+y)[f(x)+f(y)]=f(x)f(y),f(1)=2,且當(dāng)x∈(0,+∞)時(shí),f(x)>0恒成立,則下列結(jié)論正確的是(

)A.f()=6B.f(2x)=2f(x)C.f(x)為奇函數(shù)D.f(x)在區(qū)間(0,+∞)是單調(diào)遞增函數(shù)C假設(shè)存在x<0,使f(x)=0.把y用-2x代換,則有f(-x)[f(x)+f(-2x)]=f(x)f(-2x),即f(-x)f(-2x)=0,又當(dāng)x>0時(shí),f(x)>0,所以產(chǎn)生矛盾,即x<0時(shí),f(x)≠0,則f(x)≠0在函數(shù)f(x)的定義域內(nèi)恒成立.令-x代換x,y,則f(-x-x)[f(-x)+f(-x)]=f(-x)f(-x),即2f(-2x)·f(-x)=f(-x)f(-x),所以2f(-2x)=f(-x),令-x代換x,所以2f(2x)=f(x),故B錯(cuò)誤;令y=-2x,則f(x-2x)[f(x)+f(-2x)]=f(x)f(-2x),即f(-x)·[f(x)+f(-2x)]=f(x)f(-2x).化簡(jiǎn)可得f(-x)=-f(x),又函數(shù)f(x)的定義域關(guān)于原點(diǎn)對(duì)稱(chēng),所以f(x)為奇函數(shù),故C正確;令x=y=1,則f(2)[f(1)+f(1)]=f(1)f(1),解得f(2)=1,f(1)=2>f(2)=1,故D錯(cuò)誤.故選C.特殊函數(shù)模型的應(yīng)用ABD(2)(2024吉林模擬)已知函數(shù)f(x)的定義域?yàn)镽,且f(x+y)+f(x-y)=2f(x)f(y),f(0)=1,f(3x+1)=-f(-3x+1),則

f(k)=(

)A.-2 B.-1 C.0 D.1D解析(方法一)令x=0,由f(x+y)+f(x-y)=2f(x)f(y),f(0)=1,可得f(-y)=f(y),所以f(x)是偶函數(shù).因?yàn)閒(3x+1)=-f(-3x+1),所以函數(shù)f(x)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱(chēng),(方法二)由題意知函數(shù)f(x)的定義域?yàn)镽,且f(x+y)+f(x-y)=2f(x)·f(y),f(0)=1,令x=0,則f(y)+f(-y)=2f(y),即f(-y)=f(y),故f(x)為偶函數(shù);又f(3x+1)=-f(-3x+1),令x=0,則f(1)=-f(1),所以f(1)=0,又由f(3x+1)=-f(-3x+1),得f(x+1)+f(-x+1)=0,即f(x)的圖象關(guān)于點(diǎn)(1,0)成中心對(duì)稱(chēng),則f(2)=-f(0)=-1.f(x+1)+f(-x+1)=0,即f(x+2)=-f(-x),又結(jié)合f(x)為偶函數(shù),則f(x+2)=-f(x),故f(x+4)=-f(x+2)=f(x),即4為f(x)的周期,故f(3)=f(-1)=f(1)=0,f(4)=f(0)=1,故

f(k)=f(0)+[f(1)+f(2)+…+f(2

024)]=1+506×[f(1)+f(2)+f(3)+f(4)]=1+506×(0-1+0+1)=1,故選D.抽象函數(shù)性質(zhì)的綜合應(yīng)用例3(1)(多選題)(2024山東聊城一模)設(shè)f(x)是定義在R上的可導(dǎo)函數(shù),其導(dǎo)數(shù)為g(x),若f(3x+1)是奇函數(shù),且對(duì)于任意的x∈R,f(4-x)=f(x),則對(duì)于任意的k∈Z,下列說(shuō)法正確的是(

)A.4k都是g(x)的周期B.曲線(xiàn)y=g(x)關(guān)于點(diǎn)(2k,0)對(duì)稱(chēng)C.曲線(xiàn)y=g(x)關(guān)于直線(xiàn)x=2k+1對(duì)稱(chēng)D.g(x+4k)是偶函數(shù)BC解析

由f(3x+1)是奇函數(shù),故有f(3x+1)=-f(-3x+1),即有f(x+1)=-f(-x+1),故f'(x+1)=f'(-x+1),即g(x+1)=g(-x+1),故函數(shù)g(x)的圖象關(guān)于直線(xiàn)x=1對(duì)稱(chēng),由f(4-x)=f(x),則-f'(4-x)=f'(x),即-g(4-x)=g(x).故函數(shù)g(x)的圖象關(guān)于(2,0)中心對(duì)稱(chēng).由-g(4-x)=g(x),則-g(3-x)=g(x+1),又g(x+1)=g(-x+1),故g(-x+1)=-g(3-x),即有g(shù)(x+1)=-g(3+x),則g(x+3)=-g(x+5),故g(x+3)=-g(x+5)=-g(x+1),即g(x+1)=g(x+5),故g(x)=g(x+4),故g(x)的周期為4.對(duì)于A(yíng),當(dāng)k=0時(shí),4k=0,故A錯(cuò)誤;對(duì)于B,由g(x)周期為4,故g(4k-x)=g(-x),又-g(4-x)=g(x),故-g(-x)=g(x),故g(-x)=-g(x)=g(4k-x),故曲線(xiàn)y=g(x)關(guān)于點(diǎn)(2k,0)對(duì)稱(chēng),故B正確;對(duì)于C,由g(x)的周期為4,故g(4k+2-x)=g(2-x),又g(x+1)=g(-x+1),故g(x)=g(-x+2)=g(4k+2-x),故曲線(xiàn)y=g(x)關(guān)于直線(xiàn)x=2k+1對(duì)稱(chēng),故C正確;對(duì)于D,由選項(xiàng)B得-g(x)=g(4k-x),故-g(-x)=g(4k+x),又g(x)的周期為4,故有-g(-x)=-g(4k-x),故g(4k+x)=-g(4k-x),又x∈R,即g(x+4k)都是奇函數(shù),故D錯(cuò)誤.故選BC.(2)(多選題)(2024山東青島期末)已知函數(shù)f(x)及其導(dǎo)函數(shù)f'(x)的定義域均為R,若f(x)是奇函數(shù),f(2)=-f(1)≠0,且對(duì)任意x,y∈R,f(x+y)=f(x)f'(y)+f'(x)f(y),則(

)ABD解析

因?yàn)閒(x+y)=f(x)f'(y)+f'(x)f(y),令x=y=1,得f(2)=2f(1)f'(1),又因?yàn)閒(2)=-f(1)≠0,所以f'(1)=-,故A正確;因?yàn)閒(x)是定義域?yàn)镽的奇函數(shù),所以f(0)=0,且f'(x)為偶函數(shù).令y=1,可得f(x+1)=f(x)f'(1)+f'(x)f(1).①再用-x代替x,可得f(1-x)=f(-x)·f'(1)+f'(-x)f(1)=-f(x)f'(1)+f'(x)f(1),則f(x-1)=f(x)f'(1)-f'(x)f(1).②①+②,得f(x+1)+f(x-1)=2f(x)·f'(1),則f(x+1)=-f(x)-f(x-1),所以f(x+2)=-f(x+1)-f(x),f(x+3)=-f(x+2)-f(x+1)=f(x+1)+f(x)-f(x+1)=f(x),所以f(x)是周期為3的周期函數(shù),所以f(6)=f(3)=f(0)=0,故B正確;針對(duì)訓(xùn)練1.(多選題)(2024安徽安慶二模)已知定義在R上的函數(shù)f(x),滿(mǎn)足對(duì)任意的實(shí)數(shù)x,y,均有f(x+y)=f(x)+f(y)-1,且當(dāng)x>0時(shí),f(x)<1,則(

)A.f(0)=1B.f(1)+f(-1)=1C.函數(shù)f(x)為減函數(shù)D.函數(shù)y=f(x)的圖象關(guān)于點(diǎn)(0,1)對(duì)稱(chēng)ACD解析

(方法一)由f(x+y)=f(x)+f(y)-1,且當(dāng)x>0時(shí),f(x)<1,可以令f(x)=-x+1,則f(0)=1,A正確;f(1)+f(-1)=0+2=2,B錯(cuò)誤;f(x)=-x+1為減函數(shù),C正確;因?yàn)閒(0)=1,所以D正確.故選ACD.(方法二)對(duì)于A(yíng),令x=y=0,則有f(0)=f(0)+f(0)-1,故f(0)=1,故A正確;對(duì)于B,令x=1,y=-1,則有f(0)=f(1)+f(-1)-1,故f(1)+f(-1)=2,故B錯(cuò)誤;對(duì)于C,令y>0,則有f(x+y)-f(x)=f(y)-1,其中x+y>x,f(y)-1<0,令x1=x+y,x2=x,即有對(duì)?x1,x2∈R,當(dāng)x1>x2時(shí),f(x1)-f(x2)<0恒成立,即函數(shù)f(x)為減函數(shù),故C正確;對(duì)于D,令y=-x,則有f(x-x)=f(x)+f(-x)-1,又f(0)=1,故f(x)+f(-x)=2,故函數(shù)y=f(x)的圖象關(guān)于點(diǎn)(0,1)對(duì)稱(chēng),故D正確.故選ACD.D解析

由f(x+2)+f(x)=f(4),得f(x+4)+f(x+2)=f(4),則f(x+4)=f(x),即函數(shù)f(x)的周期為4.由f(2x+1)是R上的奇函數(shù),得f(-2x+1)=-f(2x+1),即f(-x+1)+f(x+1)=0,于是

D正確.故選D.3.(多選題)(2024浙江寧波期末)已知函數(shù)f(x)滿(mǎn)足:對(duì)?x,y∈R,都有f(x-y)=f(x)f(y)+f(1+x)f(1+y),且f(0)≠f(2),則以下選項(xiàng)正確的是(

)A.f(1)=0 B.f(0)=0C.f(x)+f(2-x)=0 D.f(x+4)=f(x)ACD解析

對(duì)于A(yíng),令x=y=0,則f(0)=[f(0)]2+[f(1)]2,令x=y=1,則f(0)=[f(1)]2+[f(2)]2,所以[f(0)]2=[f(2)]2,因?yàn)閒(0)≠f(2),所以f(0)=-f(2).令x=1,y=0,則f(1)=f(0)f(1)+f(1)f(2)=0,故A正確;對(duì)于B,結(jié)合選項(xiàng)A可得f(0)=[f(0)]2,所以f(0)=0,或f(0)=1.若f(0)=0,則f(0)=[f(1)]2+[f(2)]2=0,所以f(2)=0,此時(shí)與f(0)≠f(2)矛盾,舍去;若f(0)=1,則f(0)=[f(1)]2+[f(2)]2=1,解得f(2)=±1,因?yàn)閒(0)≠f(2),所以f(2)=-1,故B錯(cuò)誤;對(duì)于C,令x=0,則f(-y)=f(0)f(y)+f(1)f(1+y),因?yàn)閒(1)=0,f(0)=1,所以f(-y)=f(y),所以f(x)為偶函數(shù),令x=1,則f(1-y)=f(1)f(y)+f(2)f(1+y)=f(2)f(1+y)=-f(1+y),所以f(1-y)=-f(1+y),令y=1-x,則f(x)=-f(2-x),即f(x)+f(2-x)=0,故C正確;對(duì)于D,因?yàn)閒(x)=-f(2-x),把x換成x+2,則f(x+2)=-f(-x),又f(x)為偶函數(shù),所以f(-x)=f(x),所以f(x+2)=-f(x).把x換成x+2,則f(x+4)=-f(x+2)=f(x),所以f(x+4)=f(x),故D正確.故選ACD.4.(多選題)(2024湖南衡陽(yáng)二模)已知函數(shù)f(x),g(x)的定義域均為R,y=g(x+4)-3是奇函數(shù),且g(x)-f(x-2)=2,f(x)+g(x+6)=4,g(2)=4,則(

)A.g(4)=3 B.f(x)為奇函數(shù)C.g(x+2)為偶函數(shù)

ACD解析

由y=g(x+4)-3是奇函數(shù),則g(-x+4)-3=-g(x+4)+3,即g(-x+4)+g(x+4)=6,令x=0,則g(4)=3,故A正確;由g(x)-f(x-2)=2,g(2)=4,令x=2,則f(0)=2≠0,故f(x)不是奇函數(shù),故B錯(cuò)誤;由g(-x+4)+g(x+4)=6,把x代換成x-2,則g(-x+6)+g(x+2)=